automorphisms of uniform lattices of nilpotent lie groups...

Commun. Korean Math. Soc. 35 (2020), No. 2, pp. 653–666

https://doi.org/10.4134/CKMS.c190192

pISSN: 1225-1763 / eISSN: 2234-3024

AUTOMORPHISMS OF UNIFORM LATTICES OF

NILPOTENT LIE GROUPS UP TO DIMENSION FOUR

Jong Bum Lee and Sang Rae Lee

Reprinted from the

Communications of the Korean Mathematical Society

Vol. 35, No. 2, April 2020

c©2020 Korean Mathematical Society

Commun. Korean Math. Soc. 35 (2020), No. 2, pp. 653–666

https://doi.org/10.4134/CKMS.c190192

pISSN: 1225-1763 / eISSN: 2234-3024

AUTOMORPHISMS OF UNIFORM LATTICES OF

NILPOTENT LIE GROUPS UP TO DIMENSION FOUR

Jong Bum Lee and Sang Rae Lee

Abstract. In this paper, when G is a connected and simply connected

nilpotent Lie group of dimension less than or equal to four, we studythe uniform lattices Γ of G up to isomorphism and then we study the

structure of the automorphism group Aut(Γ) of Γ from the viewpoint ofsplitting as a natural extension.

1. Introduction

In this paper we study the group of automorphisms of any uniform lattice ofa connected and simply connected nilpotent Lie group G up to dimension four.This work was motivated by the papers [3] and [5], in which the authors consid-ered the discrete subgroup Heis(3,Z) of the three-dimensional Heisenberg groupHeis(3,R) and proved that the automorphism group Aut(Heis(3,Z)) admits asplitting as a natural extension of Z2 by GL(2,Z).

The connected and simply connected nilpotent Lie groups of dimension lessthan or equal to four are well understood. In dimension one or two, there is onlyone such Lie group, the abelian Lie group R or R2. There are two connectedand simply connected three-dimensional nilpotent Lie groups R3 and Nil3. TheLie group Nil3 is the Heisenberg group

Heis(3,R) =

1 y z

0 1 x0 0 1

| x, y, z ∈ R

.

There are three connected and simply connected four-dimensional nilpotent Liegroups R4,Nil3×R and Nil4. The Lie groups Nil3×R and Nil4 are of the form

Received June 3, 2019; Accepted September 26, 2019.

2010 Mathematics Subject Classification. Primary 22E25; Secondary 22E40.Key words and phrases. Automorphism, nilpotent Lie group, splitting, uniform lattice.The first-named author is supported in part by Basic Science Research Program through

the National Research Foundation of Korea(NRF) funded by the Ministry of Education

(NRF2016R1D1A1B01006971).

c©2020 Korean Mathematical Society

653

654 J. B. LEE AND S. R. LEE

R3 oϕ(s) R, [4], where ϕ(s) is respectively

ϕ(s) =

1 s 00 1 00 0 1

and

1 s 12s

2

0 1 s0 0 1

.

For solvable Lie groups, one may refer to [6, 7].A discrete subgroup Γ of a Lie group G is called a uniform lattice of G

if its orbit space Γ\G is compact. When G is an abelian Lie group Rn, everyuniform lattice Γ of G is isomorphic to Zn ⊂ Rn and hence Aut(Γ) ∼=Aut(Zn) ∼=GL(n,Z).

In this paper, when G is Nil3, Nil3 × R or Nil4, we first study the uniformlattices Γ of G up to isomorphism and then we study the structure of theautomorphism group Aut(Γ) of Γ from the viewpoint of splitting as a naturalextension.

2. The Lie group Nil3

The Lie group Nil3 is the Heisenberg group

Heis(3,R) =

1 y z

0 1 x0 0 1

| x, y, z ∈ R

.

We will study Aut(Γ) for any uniform lattice Γ of Nil3. The groups

Γk =

1 n `

k0 1 m0 0 1

| `,m, n ∈ Z

are uniform lattices of Nil3. Note that Γ1 = Heis(3,Z), and Γ−k ∼= Γk for allk 6= 0. It is known that any uniform lattice of Nil3 is isomorphic to exactly oneΓk for some k > 0. In [3] and [5], it is shown that the automorphism group ofΓ1 is isomorphic to the group Z2 o GL(2,Z). Utilizing the methods employedin [3] and [5], we can prove that Aut(Γk) is isomorphic to Z2 o GL(2,Z) forevery k > 0.

Lemma 2.1. The lattice Γk of Nil3 may be presented as

Γk = 〈α, β, γ | [α, β] = γ−k, [γ, α] = [γ, β] = 1 〉

with α, β and γ corresponding to the generators

α =

1 0 00 1 10 0 1

, β =

1 1 00 1 00 0 1

, γ =

1 0 1k

0 1 00 0 1

.

Because the center Z(Γk) of Γk is generated by γ, every automorphism of Γkinduces an automorphism of the quotient group Γk/Z(Γk)∼=Z2. Thus we have anatural homomorphism ϑ : Aut(Γk)→ GL(2,Z). Indeed it is well-known that

AUTOMORPHISMS OF UNIFORM LATTICES OF NILPOTENT LIE GROUPS 655

ϑ is surjective, see for example [3, Proposition 6]. The purpose of this sectionis, using the Lie algebra argument, to prove that ϑ : Aut(Γk)→ GL(2,Z) splits.

Recall that the Lie algebra nil3 of Nil3 is

nil3 =

0 b c

0 0 a0 0 0

| a, b, c ∈ R

.

Choose the canonical basis {e1, e2, e3} in nil3 where

e1 =

0 0 00 0 10 0 0

, e2 =

0 1 00 0 00 0 0

, e3 =

0 0 10 0 00 0 0

.

Then [e1, e2] = −e3 and [e3, e1] = [e3, e2] = 0. With respect to the canonicalbasis, each automorphism of nil3 has a matrix presentation.

Proposition 2.2 ([2, Proposition 2.2]). The group Aut(nil3) of all automor-phisms of the Lie algebra nil3 is isomorphic to the matrix group

a b 0c d 0u v ad− bc

| a, b, c, d, u, v ∈ R, ad− bc 6= 0

.

Since the Lie group Nil3 is simply connected, we have a canonical isomor-phism Aut(Nil3) ∼= Aut(nil3) defined by ϕ 7→ dϕ fitting in the following com-muting diagram:

nil3dϕ−−−−→ nil3xlog

yexp

Nil3ϕ−−−−→ Nil3

(2.1)

Using the diffeomorphism

exp : nil→ Nil,

0 b c0 0 a0 0 0

7→1 b c+ 1

2ab0 1 a0 0 1

we can see that if

dϕ =


,

then

ϕ = exp ◦ dϕ ◦ log :

1 y z0 1 x0 0 1

7−→1 cx+ dy z∗

0 1 ax+ by0 0 1

,(2.2)

where z∗ = (ad− bc)z + 12acx

2 + ux+ bcxy + vy + 12bdy

2. In what follows, weshall identify φ = dϕ with ϕ.


Consider the uniform lattice Γk of Nil3. Due to Mal’cev, every automorphismof Γk can be extended uniquely to a Lie group automorphism of Nil3. Thisimplies that we can regard Aut(Γk) as a subgroup of Aut(Nil3). Thus

Aut(Γk) ⊂ Aut(Nil3) = Aut(nil3)

as a subgroup of the matrix group GL(3,R). Furthermore, we have a commu-tative diagram between surjective homomorphisms

Aut(nil3)ϑ−−−−→ Aut(R2) = GL(2,R) −−−−→ 1x x

Aut(Γk)ϑ−−−−→ GL(2,Z) −−−−→ 1

where the vertical maps are inclusions, and

ϑ :


7→ (a bc d

).

Now we recall that GL(2,Z) is presented as

〈ρ, τ, κ | ρτρ = τρτ, (ρτρ)4 = 1, κρκ−1 = ρ−1, κτκ−1 = τ−1, κ2 = 1〉,where ρ, τ and κ may be taken to correspond to the matrices

ρ =

(1 10 1

), τ =

(1 0−1 1

), κ =

(−1 0

0 1

).

We seek an explicit section η of ϑ. Consider the following elements ofAut(nil3):

R =

1 1 00 1 0u1 v1 1

, T =

1 0 0−1 1 0u2 v2 1

, K =

−1 0 00 1 0u3 v3 −1

.

As observed above, these elements are regarded as elements of Aut(Nil3). As-sume that the elements R, T,K satisfy the defining relations of GL(2,Z).

RTR = TRT, (RTR)4 = I3, (RTR)2 6= I3,

K2 = I3, KRK−1 = R−1, KTK−1 = T−1.

By direct computation with (2.2) we have

R(α) = αγu1k, R(β) = αβγ(v1+1/2)k, R(γ) = γ,

T (α) = αβ−1γ(u2−1/2)k, T (β) = βγv2k, T (γ) = γ,

K(α) = α−1γu3k, K(β) = βγv3k, K(γ) = γ−1.

Using [α, γ] = [β, γ] = 1, it is immediate to check the implications

RTR(α) = TRT (α)⇒ v2 = 0,

K2(α) = α⇒ u3 = 0,


KRK−1(α) = R−1(α)⇒ u1 = 0,

KTK−1(α) = T−1(α)⇒ 2u2 + v3 = −1.

We can choose v1, u2, v3 appropriately so that R, T,K also preserve Γk. Forexample, one can take

R =

1 1 00 1 00 − 1

2 1

, T =

1 0 0−1 1 0

12 0 1

, K =

−1 0 00 1 00 −2 −1

and define η : GL(2,Z)→ Aut(nil3) by ρ 7→ R, τ 7→ T, κ 7→ K. It is straight-forward to check that the above R, T and K satisfy the remaining relations ofGL(2,Z). Therefore, the subgroup of Aut(Γk) ⊂ Aut(nil3) generated by R, Tand K is isomorphic to GL(2,Z). Consequently, η provides a desired splittingof ϑ. The splitting η is independent of Γk.

Remark 2.3. The group structure of Aut(Γk) is complicated. By consideringthe Lie algebra, we can embed Aut(Γk) as a subgroup of the matrix groupGL(3,R). This makes easy to check whether a splitting function of the surjec-tive homomorphism ϑ : Aut(Γk)→ GL(2,Z) is a “homomorphism”.

Remark 2.4 (Automorphisms of Γk). We have shown that the homomorphism

ϑ : Aut(Γk)(⊂ Aut(nil3) ⊂ GL(3,R))→ GL(2,Z)

is surjective and splits. This in particular implies that

Aut(Γk) =

H =

m11 m12 0m21 m22 0p1 p2 p3

∈ Aut(nil3)∣∣∣ H preserves Γk

.

It is now straightforward to observe further that

Aut(Γk) =

m11 m12 0m21 m22 0p1 p2 m11m22 −m21m22

∣∣∣ mij ∈ Z,pi + k

2m1im2i ∈ Z

.

Hence, the groups Aut(Γk) are identical for all k 6= 0.

In conclusion we have that

ker(ϑ) =

1 0 0

0 1 0p1 p2 1

∣∣∣ p1, p2 ∈ Z

∼= Z⊕ Z,

and

Aut(Γk) ∼= (Z⊕ Z) o GL(2,Z),

where the GL(2,Z)-action on ker(ϑ) is

g · p =(p1 p2

)(g11 g12g21 g22

)−1.(2.3)


3. The nilpotent Lie group Nil3 × R

There are three simply connected four-dimensional nilpotent Lie groups:R4,Nil3 × R and Nil4. The Lie groups Nil3 × R and Nil4 are of the formR3 oϕ(s) R, [4], where ϕ(s) is respectively

ϕ(s) =

1 s 00 1 00 0 1

and

1 s 12s

2

0 1 s0 0 1

.

The group law of R3 oϕ(s) R is

(x, s)(y, t) = (x + ϕ(s)y, s+ t),

and it can be embedded affinely in GL(4,R) as{(ϕ(s) x

0 1

)}⊂ GL(4,R),

where ϕ(s) ∈ GL(3,R) and x ∈ R3 is a column vector. Remark that Nil3 × Ris 2-step and Nil4 is 3-step. We will continue to study Aut(Γ) for a uniformlattice Γ of Nil3 × R in this section and Nil4 in the next section.

It is easy to see that any uniform lattice Γ of Nil3 × R is isomorphic tothe product Γk × Z, where Γk is a uniform lattice of Nil3, see for example[1, Corollary 6.2.5]. Hence Γ can be presented as

〈α, β, γ, δ | [α, β] = γ−k, [γ, α] = [γ, β] = [δ, α] = [δ, β] = [δ, γ] = 1 〉

with α, β, γ and δ corresponding to the generators

α =

1 0 0 00 1 0 10 0 1 00 0 0 1

, β =

1 1 0 00 1 0 00 0 1 00 0 0 1

,

γ =

1 0 0 1

k0 1 0 00 0 1 00 0 0 1

, δ =

1 0 0 00 1 0 00 0 1 10 0 0 1

.

Let us denote Γk × Z by Γk,0. Since Z(Γk,0) = 〈γ, δ〉, we obtain a canonicalsurjective homomorphism ϑ′ : Aut(Γk,0) → GL(2,Z). We will show that ϑ′

also splits. It suffices to show that Aut(Γk) can be regarded as a subgroup ofAut(Γk,0) so that the following diagram is commutative:

Aut(Γk,0)ϑ−−−−→ GL(2,Z) −−−−→ 1x x=

Aut(Γk)ϑ−−−−→ GL(2,Z) −−−−→ 1

(3.1)


Every element of Γk,0 can be written as αmβnγpδq. It can be seen easilythat

(αmβnγpδq)r = αrmβrnγrp+kr(2)mnδrq,(3.2)

βnαm = αmβnγkmn(3.3)

for all r ∈ Z. For any h ∈ Aut(Γk,0), since Z(Γk,0) = 〈γ, δ〉, we must have

h(α) = αm11βm21γp11δp21 ,

h(β) = αm12βm22γp12δp22 ,

h(γ) = γeδe′, h(δ) = γe1δe2 ,

for some integers mij , pij and e, e′, e1, e2. Since h preserves the commutatorrelations

[α, β] = γ−k, [γ, α] = [γ, β] = [δ, α] = [δ, β] = [δ, γ] = 1,

it follows from (3.2) and (3.3) that e′ = 0 and e = m11m22 −m12m21. Noticethat

(1) ϑ(h) =

(m11 m12

m21 m22

),

(2) any automorphism h of Γk can be regarded as an automorphism h ofΓk,0 by taking

p21 = p22 = e′ = e1 = 0, e2 = 1.

Consequently, we have shown that the diagram (3.1) is commutative and

since ϑ splits by Section 2 it follows that ϑ splits.

4. The nilpotent Lie group Nil4

The nilpotent Lie group Nil4 is the matrix group

Nil4 =

1 s 12s

2 x0 1 s y0 0 1 z0 0 0 1

| x, y, z, s ∈ R

and so its Lie algebra is

nil4 =

0 s 0 a0 0 s b0 0 0 c0 0 0 0

| a, b, c, s ∈ R

.

The commutator subgroup [Nil4,Nil4] is R2 (with s = z = 0), and the centerZ(Nil4) of Nil4 is R (with s = y = z = 0). Since Nil4/Z(Nil4) is a three-dimensional nilpotent Lie group, it follows that it is isomorphic to Nil3. That


is,

Nil3

π

x∼=1 −−−−→ Z(Nil4) −−−−→ Nil4 −−−−→ Nil4/Z(Nil4) −−−−→ 1

where the explicit isomorphism is given by

π :

1 s 1

2s2 ∗

0 1 s y0 0 1 z0 0 0 1

7−→1 s y

0 1 z0 0 1

.

We can see further that [Nil4,Nil4]/Z(Nil4) is isomorphic to Z(Nil3). Conse-quently, we obtain the following commutative diagram:

1 1x x1 −−−−→ Z(Nil3) −−−−→ Nil3 −−−−→ Nil3/Z(Nil3) −−−−→ 1x xπ x∼=1 −−−−→ [Nil4,Nil4] −−−−→ Nil4 −−−−→ Nil4/[Nil4,Nil4] −−−−→ 1x x

Z(Nil4)=−−−−→ Z(Nil4)x x

1 1

4.1. The lattices of Nil4

Noting that the subset of elements of Nil4 with s, x, y, z ∈ Z is not a group,we shall describe the uniform lattices of Nil4.

Let Γ be a uniform lattice of Nil4. Then Z(Nil4) ∩ Γ is a uniform lattice ofZ(Nil4) = R, and Γ/(Z(Nil4) ∩ Γ) is a uniform lattice of Nil4/Z(Nil4) = Nil3.


Hence Γ fits in the following short exact sequences:

1 1x x1 −−−−→ Z −−−−→ Γ −−−−→ Z2 −−−−→ 1x xπ x∼=1 −−−−→ Z2 −−−−→ Γ −−−−→ Z2 −−−−→ 1x x

Z =−−−−→ Zx x1 1

(4.1)

Since Γ is a uniform lattice of Nil3, it is isomorphic to some Γk. Thus wecan choose a set of generators α, β, γ, δ of Γ so that δ ∈ Z(Nil4), and α, β, γgenerate Γ. We can assume that [α, β] = γ−k with k > 0. This implies thatα, β, γ and δ may be taken to correspond to matrices in Nil4 of the form

α =

1 0 0 a0 1 0 00 0 1 10 0 0 1

, β =

1 1 1

2 b0 1 1 00 0 1 00 0 0 1

,(4.2)

γ =

1 0 0 c0 1 0 1

k0 0 1 00 0 0 1

, δ =

1 0 0 d0 1 0 00 0 1 00 0 0 1

.

A direct computation shows that the commutators among α, β, γ, δ are inde-pendent of a and b, hence we may choose a = b = 0. Moreover, [α, γ] = 1.Therefore, Γ can be presented as

〈α, β, γ, δ | [α, β] = γ−kδm, [α, γ] = 1,(4.3)

[β, γ] = δn, [α, δ] = [β, δ] = [γ, δ] = 1〉.

We choose a, b, c and d > 0 so that they satisfy

a = b = 0, dn =1

k, kc = dm+

1

2.(4.4)

Then we can assume that k, n > 0 and the choice (4.2) of matrices for α, β, γand δ with the conditions (4.4) gives rise to a realization of an abstract groupwith presentation (4.3) as a uniform lattice of Nil4. We shall denote this Γ by


Γk,m,n with k, n ∈ N and m ∈ Z. For example, we can see that

Γ1,−1,2 =

1 n2n22

2n4

20 1 n2 n30 0 1 n10 0 0 1

∣∣∣ n1, n2, n3, n4 ∈ Z

.

We remark also that Γk,m,n ∼= Γk′,m′,n′ if and only if

k = k′, n = n′, m′ = ±m mod (k, n),

see for example [1, Corollary 6.2.7].

4.2. Automorphisms of Γk,m,n

Let Γ = Γk,m,n and let h : Γ → Γ be an automorphism of Γ. Then hpreserves the diagram (4.1). This implies that the diagram (4.1) induces thefollowing commutative diagram:

Aut(Γk)ϑ−−−−→ GL(2,Z)xσ x=

Aut(Γ)Θ−−−−→ GL(2,Z)

Here σ : Aut(Γ) → Aut(Γk) is the homomorphism induced by π. We remarkalso that ϑ is surjective and split by Section 2.

Consider the elements ei ∈ nil4

e1 =

0 0 0 00 0 0 00 0 0 10 0 0 0

, e2 =

0 1 0 00 0 1 00 0 0 00 0 0 0

,

e3 =

0 0 0 m

k2n + 12k

0 0 0 1k

0 0 0 00 0 0 0

, e4 =

0 0 0 1

kn0 0 0 00 0 0 00 0 0 0

.

Then

exp e1 = α, exp e2 = β, exp e3 = γ, exp e4 = δ

generate Γ, and the nontrivial Lie brackets in nil4 between e1, e2, e3 and e4 are

[e1, e2] = −ke3 +

(m+

kn

2

)e4, [e2, e3] = ne4.(4.5)

Consider a Lie algebra automorphism φ : nil4 → nil4 of nil4. Then φ is alinear transformation of the linear space nil4 preserving all the Lie bracketsbetween the linear basis {e1, e2, e3, e4} of nil4. Because φ must preserve thelower central series of nil4, it is of the form

φ(e1) = a11e1 + a21e2 + p11e3 + p21e4,


φ(e2) = a12e1 + a22e2 + p12e3 + p22e4,

φ(e3) = a33e3 + a43e4,

φ(e4) = a44e4.

Because ϕ must preserve all the Lie brackets (4.5) including the trivial ones,we obtain that

a21 = 0,

a33 = a11a22,

a44 = a22a33 = a11a222,

a43k = a22p11n+ a11a22(a22 − 1)

(m+

kn

2

).

Consequently, we have that Aut(nil4) is isomorphic to a subgroup of the matrixgroup GL(4,R):

a11 a12 0 00 a22 0 0p11 p12 a11a22 0p21 p22 p23 a11a

222

∣∣∣ p23k = a22p11n+a11a22(a22 − 1)

(m+ kn

2

) .(4.6)

By a result of Mal’cev again, we can regard

Aut(Γ) ⊂ Aut(Nil4) = Aut(nil4)

via the following commutative diagram:

nil4φ=dϕ−−−−→ nil4xlog

yexp

Nil4ϕ−−−−→ Nil4x x

Γϕ−−−−→ Γ

(4.7)

Definition 4.1. Denote by UT(2,−) the subgroup of GL(2,−) consisting ofupper triangular matrices.

Thus we have commutative diagrams:

Aut(nil4) −−−−→ UT(2,R) −−−−→ 1x xAut(Γ)

Θ−−−−→ UT(2,Z) −−−−→ 1

(4.8)


and

Aut(Γk)ϑ−−−−→ GL(2,Z) −−−−→ 1xσ x

Aut(Γk,m,n)Θ−−−−→ UT(2,Z) −−−−→ 1.

(4.9)

This induces the following commutative diagram:

1 −−−−→ ker(ϑ) −−−−→ ϑ−1(UT(2,Z))ϑ−−−−→ UT(2,Z) −−−−→ 1x xσ x=

1 −−−−→ ker(Θ) −−−−→ Aut(Γk,m,n)Θ−−−−→ UT(2,Z) −−−−→ 1

Remark that the top extension 1→ ker(ϑ)→ ϑ−1(UT(2,Z))ϑ→ UT(2,Z)→ 1

splits by Section 2. If Θ splits then ϑ splits as well. We will study whether thebottom extension

1→ ker(Θ)→ Aut(Γk,m,n)Θ→ UT(2,Z)→ 1

splits.

4.3. Splitting of Aut(Γk,m,n)Θ→ UT(2,Z)

Remark that the subgroup UT(2,Z) of GL(2,Z) is generated by

ρ =

(1 10 1

), κ =

(−1 0

0 1

), η =

(−1 0

0 −1

).

They satisfy the relations

κ2 = η2 = 1, κη = ηκ, κρκ−1 = ρ−1, ηρη−1 = ρ.

To discuss whether Θ splits, we first lift ρ, κ and η to elements of Aut(nil4)using (4.6) as follows:

R =

1 1 0 00 1 0 0r11 r12 1 0r21 r22 r23 1

with r23k = r11n,

K =

−1 0 0 0

0 1 0 0k11 k12 −1 0k21 k22 k23 −1

with k23k = k11n,

N =

−1 0 0 0

0 −1 0 0n11 n12 1 0n21 n22 n23 −1

with n23k = −n11n− 2

(m+

kn

2

).


We first find the conditions on R,N and K to generate a subgroup ofAut(nil4) isomorphic to UT(2,Z). They must satisfy the identities

K2 = N2 = I2, KN = NK, KRK−1 = R−1, NRN−1 = R(4.10)

or equivalently they satisfy

r11 = r21 = r23 = 0, k11 = k21 = k23 = 0,

n11 = −2r12, n12 = −k12,

n23 = −2m+ kn+ nn11k

, n21 =n11n23

2, n22 =

n12n232

.

Next we find the conditions on R,K and N to preserve the lattice Γk,m,n.

For this purpose, we need to recall that a Lie algebra automorphism of nil4 isregarded as a Lie group automorphism of Nil4 via the diagram (4.7). In fact,by considering Nil4 as the Mal’cev completion of Γk,m,n, i.e., by considering

the elements of Nil4 as αr1βr2γr3δr4 for ri ∈ R, we can see that

R(α)=α, R(β)=αβγk−n11

2 δ−6m+7kn−3nn11−12r22

12 , R(γ)=γ, R(δ)=δ,

K(α)=α−1, K(β)=βγ−n12δ2k22+nn12

2 , K(γ)=γ−1, K(δ)=δ−1,

N(α)=α−1γn11δn11n23

2 , N(β)=β−1γn12δn12(n23+n)

2 , N(γ)=γδn23 , N(δ)=δ−1.

Thus R, K and N preserve the lattice Γk,m,n if and only if

n11, n12, n23 ∈ Z,n11 − k, nn12, n11n23, n12n23 ∈ 2Z,6m+ 7kn− 3nn11 ∈ 12Z,

n+ n23 = −2m+ nn11k

.

With n11 = k+ 2p (p ∈ Z) and n12 = q, n23 = −r ∈ Z, we see that the identityn+ n23 = − 2m+nn11

k reduces to

r = 2n+2(m+ pn)

k,(4.11)

the condition 6m+ 7kn− 3nn11 ∈ 12Z reduces to

3m+ 2kn− 3pn ∈ 6Z,(4.12)

and the remaining conditions reduce to the conditions

qn, rk, qr ∈ 2Z.(4.13)

By (4.11), rk is even. If n is even then by (4.11) and (4.12), m+rk−5pn ∈ 6Z,which implies that m must be even.

Consequently, we have proven the following main result.


Theorem 4.2. Given k, n ∈ N and m ∈ Z, the natural surjective homomor-phism Θ : Aut(Γk,m,n) → UT(2,Z) splits if and only if there exists an integerp such that

3m+ 2kn− 3pn ∈ 6Z,m+ pn

k∈ Z.

If n is even, then so is m.

Example 4.3. For the lattice Γ1,−1,2 of Nil4, the corresponding homomor-phism Θ cannot split because n = 2 is even and m = −1 is odd.

Consider the lattice Γ3,1,3 of Nil4. The corresponding homomorphism Θcannot split, because 3m+ 2kn− 3pn = 21− 9p /∈ 6Z.

The homomorphism Θ corresponding to the lattice Γ1,−1,3 splits becausethe above conditions are satisfied if we take p = −1.

Acknowledgement. The authors would like to thank the referee for readingthe manuscript thoroughly and correcting some grammar and typos.

References

[1] K. Dekimpe, Almost-Bieberbach groups: affine and polynomial structures, Lecture

Notes in Mathematics, 1639, Springer-Verlag, Berlin, 1996. https://doi.org/10.1007/BFb0094472

[2] K. Y. Ha and J. B. Lee, Left invariant metrics and curvatures on simply connected three-

dimensional Lie groups, Math. Nachr. 282 (2009), no. 6, 868–898. https://doi.org/10.1002/mana.200610777

[3] P. J. Kahn, Automorphisms of the discrete Heisenberg groups, preprint.

[4] J. B. Lee, K. B. Lee, J. Shin, and S. Yi, Unimodular groups of type R3oR, J. Korean Math.Soc. 44 (2007), no. 5, 1121–1137. https://doi.org/10.4134/JKMS.2007.44.5.1121

[5] D. V. Osipov, The discrete Heisenberg group and its automorphism group, Math. Notes

98 (2015), no. 1-2, 185–188; translated from Mat. Zametki 98 (2015), no. 1, 152–155.https://doi.org/10.4213/mzm10694

[6] S. V. Thuong, Metrics on 4-dimensional unimodular Lie groups, Ann. Global Anal.

Geom. 51 (2017), no. 2, 109–128. https://doi.org/10.1007/s10455-016-9527-z[7] , Classification of closed manifolds with Sol14-geometry, Geom. Dedicata 199

(2019), 373–397. https://doi.org/10.1007/s10711-018-0354-1

Jong Bum LeeDepartment of MathematicsSogang UniversitySeoul 04107, Korea

Email address: [email protected]

Sang Rae LeeDepartment of Mathematics

Texas A&M UniversityCollege Station, Texas 77843, USAEmail address: [email protected]

https://doi.org/10.1007/BFb0094472

https://doi.org/10.1007/BFb0094472

https://doi.org/10.1002/mana.200610777

https://doi.org/10.1002/mana.200610777

https://doi.org/10.4134/JKMS.2007.44.5.1121

https://doi.org/10.4213/mzm10694

https://doi.org/10.1007/s10455-016-9527-z

https://doi.org/10.1007/s10711-018-0354-1

automorphisms of uniform lattices of nilpotent lie groups...

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